Divide using synthetic division $$ ( 19x^{3}-19x^{2} ) \div ( 6 ) $$
Answer
The synthetic division table is:
$$ \begin{array}{c|rrrr}0&19&-19&0&0\\& & 0& 0& \color{black}{0} \\ \hline &\color{blue}{19}&\color{blue}{-19}&\color{blue}{0}&\color{orangered}{0} \end{array} $$The solution is:
$$ \frac{ 19x^{3}-19x^{2} }{ x } = \color{blue}{19x^{2}-19x} $$Explanation
Step 1 : Write down the coefficients of the dividend into division table.Put the zero at the left.
$$ \begin{array}{c|rrrr}\color{blue}{0}&19&-19&0&0\\& & & & \\ \hline &&&& \end{array} $$Step 1 : Bring down the leading coefficient to the bottom row.
$$ \begin{array}{c|rrrr}0&\color{orangered}{ 19 }&-19&0&0\\& & & & \\ \hline &\color{orangered}{19}&&& \end{array} $$Step 2 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 19 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&19&-19&0&0\\& & \color{blue}{0} & & \\ \hline &\color{blue}{19}&&& \end{array} $$Step 3 : Add down last column: $ \color{orangered}{ -19 } + \color{orangered}{ 0 } = \color{orangered}{ -19 } $
$$ \begin{array}{c|rrrr}0&19&\color{orangered}{ -19 }&0&0\\& & \color{orangered}{0} & & \\ \hline &19&\color{orangered}{-19}&& \end{array} $$Step 4 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ \left( -19 \right) } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&19&-19&0&0\\& & 0& \color{blue}{0} & \\ \hline &19&\color{blue}{-19}&& \end{array} $$Step 5 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&19&-19&\color{orangered}{ 0 }&0\\& & 0& \color{orangered}{0} & \\ \hline &19&-19&\color{orangered}{0}& \end{array} $$Step 6 : Multiply by the number on the left, and carry the result into the next column: $ \color{blue}{ 0 } \cdot \color{blue}{ 0 } = \color{blue}{ 0 } $.
$$ \begin{array}{c|rrrr}\color{blue}{0}&19&-19&0&0\\& & 0& 0& \color{blue}{0} \\ \hline &19&-19&\color{blue}{0}& \end{array} $$Step 7 : Add down last column: $ \color{orangered}{ 0 } + \color{orangered}{ 0 } = \color{orangered}{ 0 } $
$$ \begin{array}{c|rrrr}0&19&-19&0&\color{orangered}{ 0 }\\& & 0& 0& \color{orangered}{0} \\ \hline &\color{blue}{19}&\color{blue}{-19}&\color{blue}{0}&\color{orangered}{0} \end{array} $$Bottom line represents the quotient $ \color{blue}{ 19x^{2}-19x } $ with a remainder of $ \color{red}{ 0 } $.